BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
                                    Light of wavelength \[\lambda ,\] strikes a photoelectric surface and electrons are ejected with an energy E. Iff is to be increased to exactly twice its original value, the wavelength changes to\[\lambda ;\] where

    A) \[\lambda \]is less than \[\frac{\lambda }{2}\]   

    B) \[\lambda \]is greater than\[\frac{\lambda }{2}\]

    C)  \[\lambda \] is greater than \[\frac{\lambda }{2}\] but less than \[\lambda ,\]

    D)  \[\lambda \] is exactly equal to\[\frac{\lambda }{2}\]

    Correct Answer: C

    Solution :

    Energy of photoelectron                               \[E=\frac{hc}{\lambda }-W\]                 \[\Rightarrow \]               \[\frac{hc}{\lambda }=E+W\]                                      ?(i) where W is the work function for the metal surface  (constant).                                       \[2E=\frac{hc}{\lambda }-W\]                 \[\Rightarrow \]               \[\frac{hc}{\lambda }=2E+W\]                                   ?(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{\lambda }{\lambda }=\frac{E+W}{2E+W}\] \[\frac{\lambda }{\lambda }=\frac{(E+W)}{\left( E+\frac{W}{2} \right)}\] \[\therefore \]  \[\frac{\lambda }{\lambda }>\frac{1}{2}\] or            \[\lambda >\frac{\lambda }{2}\] \[\therefore \]  \[\lambda >\lambda >\frac{\lambda }{2}\]


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